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A 75kW car runs at a constant speed on a straight road at a speed of 20m / s to find the traction force of the car? If the power remains the same, the car needs 7500n traction when climbing,
1. From P = w / T = = f pull s / T = f pull V, f pull = P / v = 75000 / 20 = 3750n,
2, v = P / F = 75000 / 7500 = 10m / s
What is the traction force of a 45KW car driving at a speed of 54km / h for 20km on a flat road?
Speed unchanged, straight road
F traction = resistance
P=FV
F=45000/15 (54km/h=15m/s)
=3000N
A car with a power of 29.4 kW has driven 20 km on a straight road at a speed of 15 m / s. if the engine efficiency of the car is 30%
How many kilograms of gasoline will be consumed through this journey (the calorific value of gasoline is 4.26 * 10 times of coke / kg)
Automobile energy consumption q = Pt = PS / v = 29400 × 20000 / 15 = 3.92 × 10 ^ 7 (J) effective energy provided by mkg gasoline q = m oil Q oil η = m oil × 4.26 × 10 ^ 7 × 0.3 = 1.278 × 10 ^ 7 (J)
The energy of the above two formulas is equal, so m oil = 3.067kg
A: it takes 3.067 kg of gasoline to get through this journey
The resistance of a bicycle running at normal speed on a straight road is about 0.02 times of the total weight of the vehicle and the person, and the power of the cyclist is closest to how many watts
1 W 100 W 1 kW · 10 kW
Choose B
When the body weight is about 600N, the resistance level is 12n
P = FV, the speed V of bicycle is less than 10
Then p is about 100W
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